Question

two equal loads of 2500N are suspended by a string abcd at point B and D find tension in the string AB BC CD of the string​

Answer 1

Answer:

TAB = 4330 N

TBC = 2500 N

TCD = 2500 N

Explanation:

CORRECT ANSWER

Answer 2

First string ABCD is split in to two parts and consider the joints B and C separately

Let $T_1$ = Tension in String AB.

$T_2$= Tension in String BC

$T_3$ = Tension in String CD

Since at joint B there are three forces are acting.

So Apply Lamis theorem at joint B

$\frac{T_1}{sin60^o}=\frac{T_2}{sin150^o}=\frac{2500}{sin150^o}$

°

$T_1$ = $\frac{sin60^o X 2500}{sin150^o}$

= 4330 N

$T_2$ = $\frac{sin150^oX2500}{sin150^o}$= 2500N

Again Apply Lamis theorem at joint C

$\frac{T_2}{sin120^o}=\frac{T_3}{sin120^o}$

$T_3=\frac{sin120^oX2500}{sin120^o}$

$T_3=2500N$

∴ $T_1=4330 N\\T_2=2500N\\T_3=2500N$

(Refer the attachment for the diagram)